Q.28. A mass of 1 kg kept in a inclined plane with 30° inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. Th

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Q.28. A mass of 1 kg kept in a inclined plane with 30° inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2s is J. (Take g = 10 m/s²)

FirstInTest · Accepted Answer

Correct answer: A. Sol. Mass m = 1kg. Angle of inclination = 30°. Velocity v = 4 m/s (constant). Time t = 2s. g = 10 m/s².
Work done by friction = $f_k \cdot d$. Since the assembly moves up with constant velocity, the net force is zero. The frictional force is $f_k = \mu_k N = \mu_k mg \cos	heta$. However, the problem states the *whole assembly* is moved up with constant velocity. This implies the block is still at rest relative to the inclined plane, and the work done by friction is due to the motion of the plane itself. The solution provided calculates work done as $W = mg \sin	heta \cdot d$. The distance moved is $d = v \cdot t = 4 \cdot 2 = 8m$. So, $W = 1 \cdot 10 \cdot \sin(30°) \cdot 8 = 1 \cdot 10 \cdot (1/2) \cdot 8 = 40 J$. This is the work done against gravity. The solution provided calculates $W = mg \sin 30° = 5N$ as the force, and then $W = 4 	imes 2 = 8 J$ for work done by friction. This seems to be a misinterpretation of the solution text. Let's re-evaluate based on the solution's calculation: $f = mg \sin 30° = 1 	imes 10 	imes 0.5 = 5 N$. Work done in 2 sec will be $W = f 	imes d = 5 N 	imes (4 m/s 	imes 2 s) = 5 N 	imes 8 m = 40 J$. The provided solution has $W = mg \sin 30° = 5N$ and then $W = 4 	imes 2 = 8 J$. This is inconsistent. Let's look at the options. If the question implies work done by friction *on the block* relative to the ground, and the block is at rest on the plane, then the friction force is static friction, $f_s = mg \sin	heta = 5N$. If the plane moves up by $d = 8m$, the work done by static friction on the block is $W = f_s d = 5 	imes 8 = 40 J$. If the question implies the work done by the force that moves the assembly, it would be $40 J$. The given answer is (1) 20. Let's re-examine the solution's steps: $f = mg \sin 30° = 5N$. Work done in 2 sec will be $d = 4 	imes 2 = 8m$. $W = (5)(8) \cos 60° = 20 J$. This implies the force of friction is $5N$, and the angle between force and displacement is $60°$. This is unusual. If the block is at rest on the plane, and the plane moves up, the friction force is parallel to the plane, and displacement is also parallel to the plane. So angle should be 0 or 180. If the work done by friction is $W = 20 J$, then it must be $f_k d \cos\phi = 5 	imes 8 	imes \cos\phi = 40 \cos\phi = 20$, so $\cos\phi = 1/2$, meaning $\phi = 60°$. This angle is not naturally derived from the problem statement. Given the options and the provided answer, there might be a specific interpretation or a missing piece of information. Assuming the solution's calculation $W = (5)(8) \cos 60° = 20 J$ leads to option (1).

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